Exponential functions 35
Hence α =
1
θ
ln
R
R 0
=
1
1500
ln
6 × 10 3
5 × 10 3
=
1
1500
(0.1823215 ...)
= 1.215477 · · ·× 10 −4
Hence α = 1.215 × 10 −4 ,
correct to 4 significant figures.
From above, ln
R
R 0
= αθ
hence
θ =
1
α
ln
R
R 0
When R = 5.4 × 10 3 , α= 1.215477 ... × 10 −4 and
R 0 = 5 ×10 3
θ =
1
1.215477 ... × 10 −4 ln
5.4 × 10 3
5 × 10 3
=
10 4
1.215477 ...
(7.696104 ... × 10
−2
)
= 633
◦ C, correct to the nearest degree.
Problem 18. In an experiment involving
Newton’s law of cooling, the temperature θ( ◦ C) is
given by θ = θ 0 e −kt . Find the value of constant k
when θ 0 = 56.6 ◦ C, θ = 16.5 ◦ C and t = 83.0 seconds.
Transposing
θ = θ 0 e −kt gives
θ
θ 0
= e −kt
from which
θ 0
θ
=
1
e −kt = e kt
Taking Napierian logarithms of both sides gives:
ln
θ 0
θ
= kt
from which,
k =
1
t
ln
θ 0
θ
=
1
83.0
ln
56.6
16.5
=
1
83.0
(1.2326486 ...)
Hence k = 1.485 × 10 −2
Problem 19. The current i amperes flowing in a
capacitor at time t seconds is given by
i = 8.0(1 − e
−t
CR ), where the circuit resistance R is
25 ×10 3 ohms and capacitance C is
16 ×10 −6 farads. Determine (a) the current i after
0.5 seconds and (b) the time, to the nearest
millisecond, for the current to reach 6.0 A. Sketch
the graph of current against time.
(a) Current i = 8.0(1 − e
−t
CR )
= 8.0[1 − e
−0.5
(16 ×10 −6 )(25 ×10 3 ) ] =8.0(1 − e −1.25 )
= 8.0(1 − 0.2865047 ...)= 8.0(0.7134952 ...)
= 5.71 amperes
(b) Transposing i = 8.0(1 − e
−t
CR )
gives
i
8.0
= 1 −e
−t
CR
from which, e
−t
CR = 1 −
i
8.0
=
8.0 − i
8.0
Taking the reciprocal of both sides gives:
e
t
CR =
8.0
8.0 − i
Taking Napierian logarithms of both sides gives:
t
CR
= ln
8.0
8.0 − i
Hence
t = CRln
8.0
8.0 − i
= (16 × 10
−6
)(25 × 10
3
) ln
8.0
8.0 − 6.0
when i = 6.0 amperes,
i.e. t =
400
10 3 ln
8.0
2.0
= 0.4 ln4.0
= 0.4(1.3862943 ...) = 0.5545 s
= 555 ms, to the nearest millisecond.
A graph of current against time is shown in Fig. 4.6.